SGU 548 Dragons and Princesses

来源：互联网发布：ods直播软件编辑：程序博客网时间：2024/05/19 13:42

题意：

n个格子每个格子有龙或者公主勇士从1走到n 路过龙可以杀死可以不杀杀死有钱拿路过公主如果之前杀龙的数量满足公主要求就会停止行走问勇士想多拿钱但是必须要满足n格子的公主最多拿多少钱

思路：

公主只限制杀龙的数量因此不想停下来结婚就控制杀龙的数量即可如果要放弃一些龙那么一定会贪心放弃钱少的龙最后判断一下能不能和n格子的公主结婚即可

代码：

#include<cstdio>#include<iostream>#include<cstring>#include<string>#include<algorithm>#include<map>#include<set>#include<vector>#include<queue>#include<cstdlib>#include<ctime>#include<cmath>using namespace std;#define N 200010int n;struct dragon {int x, id;bool operator<(const dragon ff) const {return x > ff.x;}} now;priority_queue<dragon> q;int ans[N], tot, sum;int main() {int i, k, s;char who[100];while (!q.empty())q.pop();scanf("%d", &n);for (i = 2; i <= n; i++) {scanf("%s%d", who, &k);if (i == n)break;if (who[0] == 'd') {now.x = k;now.id = i;q.push(now);} else {s = q.size();if (s >= k) {k = s - k + 1;while (k--)q.pop();}}}s = q.size();if (s < k)printf("-1\n");else {tot = sum = 0;while (!q.empty()) {now = q.top();q.pop();sum += now.x;ans[tot++] = now.id;}printf("%d\n", sum);printf("%d\n", tot);sort(ans, ans + tot);for (i = 0; i < tot; i++)printf("%d%s", ans[i], (i == tot - 1) ? "\n" : " ");}return 0;}

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